Question

In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 4.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as

θ = 2.30t² − 0.850t³

where θ is in radians and t is in seconds.

(a) Find the maximum angular speed of the roller.
rad/s

(b) What is the maximum tangential speed of a point on the rim of the roller?
m/s

(c) At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation?
s

(d) Through how many rotations has the roller turned between t = 0 and the time found in part (c)?

rotations

Answer 1

w = dtheta/dt

theta = 2.30t^2 - 0.850t^3

w = 4.6t - 2.55t^2

it will be maximum

dw/dt = 4.6- 5.1t = 0

t = 4.6/5.10 = 0.9 s

w_max at 0.9 s

w_max = 4.6 * 0.9 - 2.55 * 0.9^2

w_max = 2.0745 rad/s

part b )

vmax = w_max *r

r = diameter/2 = 4/2 = 2 m

vmax = 4.149 m/s

part c )

The roller reverses its direction when the angular velocity is zero

w = 4.6t - 2.55t^2 = 0

t = 4.6/2.55

The driving force should be removed from the roller at t = 1.8 s

part d )

put t = 1.8s in theta

theta = 2.3 x 1.8^2 - 0.850*1.8^3

theta = 2.4948 rad

2.4948 * 1rotation/2pi rad = 0.397 rotation